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Aleonysh [2.5K]
3 years ago
8

A consumer group wants to know if an automobile insurance company with thousands of customers has an average insurance payout fo

r all their customers that is greater than $500 per insurance claim. They know that most customers have zero payouts and a few have substantial payouts. The consumer group collects a random sample of 18 customers and computes a mean payout per claim of $579.80 with a standard deviation of $751.30.
Is it appropriate for the consumer group to perform a hypothesis test for the mean payout of all customers?
A. Yes, it is appropriate because the population standard deviation is unknown.
B. Yes, it is appropriate because the sample size is large enough, so the condition that the sampling distribution of the sample mean be approximately normal is satisfied.
C. No, it is not appropriate because the sample is more than 10 percent of the population, so a condition for independence is not satisfied.
D. No, it is not appropriate because the standard deviation is greater than the mean payout, so the condition that the sampling distribution of the sample mean be approximately normal is not satisfied.
E. No, it is not appropriate because the distribution of the population is skewed and the sample size is not large enough to satisfy the condition that the sampling distribution of the sample mean be approximately normal.
Mathematics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

E. No, it is not appropriate because the distribution of the population is skewed and the sample size is not large enough to satisfy the condition that the sampling distribution of the sample mean be approximately normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

Standard deviation larger than the sample mean means that the distribution is skewed.

By the Central Limit Theorem, when the distribution is skewed, normality is assumed for samples sizes of 30 or higher. In this question, the sample is of 18, which is less than 30, so the hypothesis test is not appropriated, and the correct answer is given by option E.

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Michelle is 7 years older than her sister Joan, and Joan is 3 years younger than their brother Ryan. If the sum of their ages is
Zielflug [23.3K]

Answer:

(C) 18

Step-by-step explanation:

We can create a systems of equations. Assuming m is Michelle's age, j is Joan's age, and r is Ryan's age, the equations are:

m = j + 7

j = r-3

m+j+r = 64

We can use substitution, since we know the "values" of m and j.

(j+7)+(r-3)+r = 64\\(j+7)+(2r-3)=64\\2r + j + 4 = 64\\2r + j = 60\\\\

r = 21, j = 18

So we know that Joan is 18 years old.

Hope this helped!

8 0
3 years ago
Brayden has 342 marbles in bags. If there are 9 marbles in each bag, how many bags of marbles does Brayden have If Brayden gives
Firlakuza [10]
Brayden has 38 bags.
After giving away 15 bags of marbles, Brayden is left with 23 bags.

To find out how many bags he has we do 342 divided by 9 which is 38.

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We then do 342 - 135 to find out how many marbles he has left which is 207 and divide 207 by 9 to convert that into bags, which is 23.
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Answer: I would say rolling a number cube

Step-by-step explanation:

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3 years ago
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THE answer I need some help i do not get it
mamaluj [8]
Hi, AlaynaSanjurjo!
All we need to do is multiply 93 by 3, then divide it by 9. Since we have 3 boxes, and 93 tangerines in each box, we have to add them all up and divide them equally. I am sure you are capable of doing the equation now.
93 * 3 / 9

I hope I helped!

~Olivia

(P.S. I am working for Virtuoso and many don't consider giving others brainliest answers, but could you give me one? I need two more!)
6 0
3 years ago
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